Question

The equation $\left|\frac{z-z_1}{z-z_2}\right|=k$ represents a circle if

• A. $k=1$
• B. $k\ne 1$
• C. $k\ne 1$ and $k<0$
• D. $kϵR$

Answer 1

The correct option is B $k\ne 1$

Let $z=x+iy$
$z_1=x_1+i{y}_1$
$z_2=x_2+i{y}_2$
Now,
$\left|\frac{(x-x_1)+i(y-y_1)}{(x-x_2)+i(y-y_2)}\right|=k$
$\Rightarrow (x-x_1{)}^2+(y-y_1{)}^2=k^2\left\{\right(x-x_2{)}^2+(y-y_2{)}^2\}$
$\Rightarrow x^2(1-k^2)+y^2(1-k^2)-2x{x}_1-2y{y}_1+x{{}_1}^2-x{{}_2}^2{k}^2+y{{}_1}^2-y{{}_2}^2{k}^2+2k^{2}x{x}_2+2k^{2}y{y}_2$
$\Rightarrow x^2(1-k^2)+y^2(1-k^2)-2x(x_1-k^2{x}_2)-2y(y_1-k^2{y}_2)+x{{}_1}^2+y{{}_1}^2-k^{2}x{{}_2}^2-k^{2}y{{}_2}^2=0$
For circle, $1-k^2\ne 0$
$k^2\ne 1$ {k is positive}
$k\ne 1$